# Thread: compound interest problem(#of yrs): through log property.

1. ## compound interest problem(#of yrs): through log property.

ok guys this is really bugging me. any help would be greatly appreciated.
p-3500
a-4000
r-.0525
quarterly rate.so n=4
so the question is asking for the number of years. so t is my unknown variable.

i set it up like this.

4000= 3500 (1+.O525/4)^4t

so far so good ryt?
4000=3545.9375 ^4t

using the property of log that can be written as

log 4000= 4t log 3545.9375

so ill get
log 4000/log 3545.9375=4t

1.01473=4t
.2536=t

my question is how come im not getting 2.5 something years. i need to multiply it by ten to get the real answer?
is it because the base of log is 10 or am i doing something wrong.
thanks

im working on a 2nd problem
this one im totally off
a=10,000 p=5000 r=.09 n=2

10000=5000(1+.09/2)^2t
10000=5225^2t
log10000=2t log 5225
log10000/log5225=2t
.5379=t

7.9

2. Originally Posted by lutwey
ok guys this is really bugging me. any help would be greatly appreciated.
p-3500
a-4000
r-.0525
quarterly rate.so n=4
so the question is asking for the number of years. so t is my unknown variable.

i set it up like this.

4000= 3500 (1+.O525/4)^4t

so far so good ryt? yes, to the above point

4000=3545.9375 ^4t this step is incorrect ... divide by 3500 first
...

3. thanks for the quick response.
you cal always multiply them ryt?

4. Originally Posted by lutwey
thanks for the quick response.
you cal always multiply them ryt?
order of operations ...

if you had $5 \cdot 3^3$ , would that equal $15^3$ ???

5. Hi lutwey,
second problem
10000=5000(1+.09/2)^2t
Simplify
2=(1+.045)^2t
log 2= 2t log 1.045
2t=log2/log1.045 =15.75
t=7.87

bjh

6. aahhhhhh!!! hahahahaha

thanks skeeter and bj hopper.
now i got it, been away from math for a couple of yrs.

thanks a lot.